What is the tangent of f(x)=x^2+5,\at x=2 ?

The tangent of f(x)=x^2+5,\at x=2 is y=4x+1

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tangent of f(x)=x^2+5,\at x=2

tangent of

f (x) = x^{2} + 5, at x = 2

y = 4 x + 1

Solution steps

Find the tangent point:

(2, 9)

Find the slope of

f (x) = x^{2} + 5 : \frac{df ( x )}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4

Find the line with slope m=

4

and passing through

(2, 9) : y = 4 x + 1

y = 4 x + 1

y = x^{3} - 4 x, (2, 0)

y^{^{''}} + 25 y = - 30 sin (5 t)

\int_{2}^{4} \frac{1}{( x - 1 ) ( x - 4 )} d x

\frac{d}{d x} (2 (1 + cos (x)))

\frac{d}{d x} (\frac{1 - e ^{7 x}}{1 + e ^{7 x}})

What is the tangent of f(x)=x^2+5,\at x=2 ?
The tangent of f(x)=x^2+5,\at x=2 is y=4x+1